consider g(x)= {a sin x + b, if x ≤ 2pi} {x^2 - pi x + 2, if x > 2pi}
A. Find the values of a a b such that g(x) is a differentiable function.
B. Write the equation of the tangent line to g(x) at x = 2pi.
C. Use the tangent line equation from part B to write an approximation for the value of g(6). Do not simplify.
sayyyyyyyyyyyy that
f(x) = a sinx + b
h(x) = x^2 - πx + 2
find h(2pi) and set it equal to f(2pi). also find h'(2pi) and set it equal to f'(2pi). together that should give you enough info to find a and b, and then the rest of the problem tests other calc concepts.
Part a:
g(x) = { asinx + b, for x ≤ 2π
.........{ x² - πx + 2, for x > 2π
g must be continuous such that
lim x→2π⁻ g(x) = lim x→2π⁺ g(x)
Lefthand limit:
lim x→2π⁻ (asinx + b) = asin(2π) + b = 0 + b = b
Right hand limit:
lim x→2π⁺ (x² - πx + 2) =
(2π)² - π(2π) + 2
4π² - 2π² + 2
2π² + 2
b = 2π² + 2
Also, for the derivative to exist at all values of x you must also check the limit of the derivative to see that the derivative approaches the same value from both sides:
..........{ acosx, for x < 2π
g'(x) = { ?, for x = 2π
..........{ 2x - π, for x > 2π
lim x→2π⁻ g'(x) =
lim x→2π⁻ (acosx) = acos(2π) = a
lim x→2π⁺ g'(x) =
lim x→2π⁺ (2x - π) = 2(2π) - π = 3π
a = 3π
So,
g(x) = { 3πsinx + 2π² + 2, for x ≤ 2π
.........{ x² - πx + 2, for x > 2π,
and g'(2π) = 3π
Part b:
g(2π) = b = 2π² + 2
This gives you the point (2π, 2π² + 2)
y - (2π² + 2) = 3π(x - 2π)
y - 2π² - 2 = 2πx - 6π²
y = 2πx - 4π² + 2
Part c:
g(6) ≈ g(2π) + g'(π)(6 - 2π)
g(6) ≈ (2π² + 2) + 3π(6 - 2π)
g(6) ≈ 2π² + 2 + 18π - 6π²
g(6) ≈ -4π² + 18π + 2
g(6) ≈ 19.07
Using a calculator, I evaluated g(6) and got 19.11
A. To ensure that g(x) is differentiable, we need the two parts of the function to "meet" smoothly at x = 2π. This means that the two parts must have the same value and slope at x = 2π.
Let's find the value of a and b by equating the two parts of g(x) at x = 2π:
a sin(2π) + b = (2π)^2 - π(2π) + 2
Since sin(2π) = 0, the equation becomes:
b = 4π^2 - 2π^2 + 2
Simplifying, we have:
b = 2π^2 + 2
Now, let's consider the slope of g(x) on each side of x = 2π. The derivative of the first part of g(x) is a cos(x), and for the second part, the derivative is 2x - π.
Equating the slopes at x = 2π:
a cos(2π) = 2(2π) - π
Since cos(2π) = 1, the equation becomes:
a = 3π
Therefore, the values of a and b that make g(x) a differentiable function are a = 3π and b = 2π^2 + 2.
B. To find the equation of the tangent line to g(x) at x = 2π, we need to find the slope of the tangent line and a point on the line.
The slope of the tangent line is equal to the derivative of g(x) at x = 2π. Taking the derivative:
g'(x) = a cos(x) for x ≤ 2π (since b is a constant)
g'(x) = 2x - π for x > 2π (since a is a constant)
Plugging in x = 2π into the second part of g'(x), we have:
g'(2π) = 2(2π) - π = 4π - π = 3π
So, the slope of the tangent line is 3π.
To find a point on the line, we can plug x = 2π into g(x):
g(2π) = a sin(2π) + b
= 0 + (2π)^2 - π(2π) + 2 (using the given values of a and b)
= 4π^2 - 2π^2 + 2
= 2π^2 + 2
So, the point on the line is (2π, 2π^2 + 2).
Therefore, the equation of the tangent line to g(x) at x = 2π is:
y - (2π^2 + 2) = 3π(x - 2π)
C. We can use the tangent line equation from part B as an approximation for the value of g(6). To do this, we substitute x = 6 into the equation of the tangent line:
y - (2π^2 + 2) = 3π(6 - 2π)
This gives us an equation that approximates the value of g(6).
since each part of g is differentiable, we need the slopes to match up at x=2π, so that dg/dx is defined everywhere.
dg/dx =
a cos x for x <= 2π
2x-π for x > 2π
so, we need
a cos(2π) = 2(2π)-π
a = 3π
g(x) is thus
3π sinx + b
x^2 - πx + 2
and we need g to be continuous, so needs must
3π*0 + b = (2π)^2 - π(2π) + 2
b = 2π^2 + 2
g(x) =
3π sinx + 2π^2 + 2
x^2 - πx + 2
g(2π) = 2π^2+2
g'(2π) = 3π
so we want the line through (2π,2π^2+2) with slope 3π:
y-(2π^2+2) = 3π(x-2π)
when x=6,
y-(2π^2+2) = 3π(6-2π)
y = 18π-6π^2 + 2π^2+2
= 18π-4π^2+2 = 19.070
In fact, g(6) = 3π sin(6) + 2π^2 + 2 = 19.106